Simplify and expand the following expression: $ \dfrac{x + 6}{4x + 1}-\dfrac{4x - 6}{x - 3} $
In order to subtract expressions, they must have a common denominator. Get both fractions over a common denominator of $(4x + 1)(x - 3)$ Multiply the first term by $\dfrac{x - 3}{x - 3}$ $ \begin{align*} \dfrac{x + 6}{4x + 1} \times \dfrac{x - 3}{x - 3} & = \dfrac{(x + 6)(x - 3)}{(4x + 1)(x - 3)} \\ & = \dfrac{x^2 + 3x - 18}{(4x + 1)(x - 3)}\end{align*} $ Multiply the second term by $\dfrac{4x + 1}{4x + 1}$ $ \begin{align*} \dfrac{4x - 6}{x - 3} \times \dfrac{4x + 1}{4x + 1} & = \dfrac{(4x - 6)(4x + 1)}{(x - 3)(4x + 1)} \\ & = \dfrac{16x^2 - 20x - 6}{(x - 3)(4x + 1)}\end{align*} $ Now we have: $ = \dfrac{x^2 + 3x - 18}{(4x + 1)(x - 3)} - \dfrac{16x^2 - 20x - 6}{(x - 3)(4x + 1)} $ Now both terms have a common denominator we can subtract the numerators: $ = \dfrac{x^2 + 3x - 18 - (16x^2 - 20x - 6)}{(4x + 1)(x - 3)} $ $ = \dfrac{x^2 + 3x - 18 - 16x^2 + 20x + 6}{(4x + 1)(x - 3)} $ $ = \dfrac{-15x^2 + 23x - 12}{(4x + 1)(x - 3)}$ Expand the denominator: $ = \dfrac{-15x^2 + 23x - 12}{4x^2 - 11x - 3}$